Physics problem

**jasonlewiz** · November 21st 2009, 05:50 AM

A cross country skier skis is 2.80 km in the direction 45.0˚ west of south, then 7.40 km in the direction 30.0˚ north of east, and finally 3.30 km in the direction 22.0˚ south of west. A) Show these displacements in a diagram. B) How far is the skier from the starting point?

**skeeter** · November 21st 2009, 06:24 AM

Originally Posted by jasonlewiz

A cross country skier skis is 2.80 km in the direction 45.0˚ west of south, then 7.40 km in the direction 30.0˚ north of east, and finally 3.30 km in the direction 22.0˚ south of west. A) Show these displacements in a diagram. B) How far is the skier from the starting point?

let the skier start at the origin. use components to find the displacement in the x and y directions ...

$r_x = -2.8\cos(45) + 7.4\cos(30) - 3.3\cos(22)$

$r_y = -2.8\sin(45) + 7.4\sin(30) - 3.3\sin(22)$

$|r| = \sqrt{(r_x)^2 + (r_y)^2}$

**aidan** · November 21st 2009, 07:04 AM

Originally Posted by skeeter

let the skier start at the origin. use components to find the displacement in the x and y directions ...

$r_x = -2.8\cos(45) + 7.4\cos(30) - 3.3\cos(22)$

$r_y = -2.8\sin(45) + 7.4\cos(30) - 3.3\cos(22)$

$|r| = \sqrt{(r_x)^2 + (r_y)^2}$

7.4\cos(30) for both x&y displacements & -3.3\cos(22) for both x&y displacements

I'll bet you did that just to see if we were awake?

.

**skeeter** · November 21st 2009, 07:07 AM

no ... I'm not awake yet

$r_x = -2.8\cos(45) + 7.4\cos(30) - 3.3\cos(22)$

$r_y = -2.8\sin(45) + 7.4\sin(30) - 3.3\sin(22)$

$|r| = \sqrt{(r_x)^2 + (r_y)^2}$

copied and only changed one , sorry.

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